3.73 \(\int \frac {\sinh ^3(a+b x^n)}{x^2} \, dx\)

Optimal. Leaf size=154 \[ -\frac {e^{3 a} 3^{\frac {1}{n}} \left (-b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-3 b x^n\right )}{8 n x}+\frac {3 e^a \left (-b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-b x^n\right )}{8 n x}-\frac {3 e^{-a} \left (b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},b x^n\right )}{8 n x}+\frac {e^{-3 a} 3^{\frac {1}{n}} \left (b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},3 b x^n\right )}{8 n x} \]

[Out]

-1/8*3^(1/n)*exp(3*a)*(-b*x^n)^(1/n)*GAMMA(-1/n,-3*b*x^n)/n/x+3/8*exp(a)*(-b*x^n)^(1/n)*GAMMA(-1/n,-b*x^n)/n/x
-3/8*(b*x^n)^(1/n)*GAMMA(-1/n,b*x^n)/exp(a)/n/x+1/8*3^(1/n)*(b*x^n)^(1/n)*GAMMA(-1/n,3*b*x^n)/exp(3*a)/n/x

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Rubi [A]  time = 0.18, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5362, 5360, 2218} \[ -\frac {e^{3 a} 3^{\frac {1}{n}} \left (-b x^n\right )^{\frac {1}{n}} \text {Gamma}\left (-\frac {1}{n},-3 b x^n\right )}{8 n x}+\frac {3 e^a \left (-b x^n\right )^{\frac {1}{n}} \text {Gamma}\left (-\frac {1}{n},-b x^n\right )}{8 n x}-\frac {3 e^{-a} \left (b x^n\right )^{\frac {1}{n}} \text {Gamma}\left (-\frac {1}{n},b x^n\right )}{8 n x}+\frac {e^{-3 a} 3^{\frac {1}{n}} \left (b x^n\right )^{\frac {1}{n}} \text {Gamma}\left (-\frac {1}{n},3 b x^n\right )}{8 n x} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x^n]^3/x^2,x]

[Out]

-(3^n^(-1)*E^(3*a)*(-(b*x^n))^n^(-1)*Gamma[-n^(-1), -3*b*x^n])/(8*n*x) + (3*E^a*(-(b*x^n))^n^(-1)*Gamma[-n^(-1
), -(b*x^n)])/(8*n*x) - (3*(b*x^n)^n^(-1)*Gamma[-n^(-1), b*x^n])/(8*E^a*n*x) + (3^n^(-1)*(b*x^n)^n^(-1)*Gamma[
-n^(-1), 3*b*x^n])/(8*E^(3*a)*n*x)

Rule 2218

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Simp[(F^a*(e + f*
x)^(m + 1)*Gamma[(m + 1)/n, -(b*(c + d*x)^n*Log[F])])/(f*n*(-(b*(c + d*x)^n*Log[F]))^((m + 1)/n)), x] /; FreeQ
[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rule 5360

Int[((e_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Dist[1/2, Int[(e*x)^m*E^(c + d*x^n), x], x]
- Dist[1/2, Int[(e*x)^m*E^(-c - d*x^n), x], x] /; FreeQ[{c, d, e, m, n}, x]

Rule 5362

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(
e*x)^m, (a + b*Sinh[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\sinh ^3\left (a+b x^n\right )}{x^2} \, dx &=\int \left (-\frac {3 \sinh \left (a+b x^n\right )}{4 x^2}+\frac {\sinh \left (3 a+3 b x^n\right )}{4 x^2}\right ) \, dx\\ &=\frac {1}{4} \int \frac {\sinh \left (3 a+3 b x^n\right )}{x^2} \, dx-\frac {3}{4} \int \frac {\sinh \left (a+b x^n\right )}{x^2} \, dx\\ &=-\left (\frac {1}{8} \int \frac {e^{-3 a-3 b x^n}}{x^2} \, dx\right )+\frac {1}{8} \int \frac {e^{3 a+3 b x^n}}{x^2} \, dx+\frac {3}{8} \int \frac {e^{-a-b x^n}}{x^2} \, dx-\frac {3}{8} \int \frac {e^{a+b x^n}}{x^2} \, dx\\ &=-\frac {3^{\frac {1}{n}} e^{3 a} \left (-b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-3 b x^n\right )}{8 n x}+\frac {3 e^a \left (-b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-b x^n\right )}{8 n x}-\frac {3 e^{-a} \left (b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},b x^n\right )}{8 n x}+\frac {3^{\frac {1}{n}} e^{-3 a} \left (b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},3 b x^n\right )}{8 n x}\\ \end {align*}

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Mathematica [A]  time = 1.41, size = 126, normalized size = 0.82 \[ \frac {e^{-3 a} \left (e^{6 a} \left (-3^{\frac {1}{n}}\right ) \left (-b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-3 b x^n\right )+3 e^{4 a} \left (-b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-b x^n\right )+\left (b x^n\right )^{\frac {1}{n}} \left (3^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},3 b x^n\right )-3 e^{2 a} \Gamma \left (-\frac {1}{n},b x^n\right )\right )\right )}{8 n x} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x^n]^3/x^2,x]

[Out]

(-(3^n^(-1)*E^(6*a)*(-(b*x^n))^n^(-1)*Gamma[-n^(-1), -3*b*x^n]) + 3*E^(4*a)*(-(b*x^n))^n^(-1)*Gamma[-n^(-1), -
(b*x^n)] + (b*x^n)^n^(-1)*(-3*E^(2*a)*Gamma[-n^(-1), b*x^n] + 3^n^(-1)*Gamma[-n^(-1), 3*b*x^n]))/(8*E^(3*a)*n*
x)

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sinh \left (b x^{n} + a\right )^{3}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b*x^n)^3/x^2,x, algorithm="fricas")

[Out]

integral(sinh(b*x^n + a)^3/x^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (b x^{n} + a\right )^{3}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b*x^n)^3/x^2,x, algorithm="giac")

[Out]

integrate(sinh(b*x^n + a)^3/x^2, x)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{3}\left (a +b \,x^{n}\right )}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a+b*x^n)^3/x^2,x)

[Out]

int(sinh(a+b*x^n)^3/x^2,x)

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maxima [A]  time = 0.46, size = 133, normalized size = 0.86 \[ \frac {\left (3 \, b x^{n}\right )^{\left (\frac {1}{n}\right )} e^{\left (-3 \, a\right )} \Gamma \left (-\frac {1}{n}, 3 \, b x^{n}\right )}{8 \, n x} - \frac {3 \, \left (b x^{n}\right )^{\left (\frac {1}{n}\right )} e^{\left (-a\right )} \Gamma \left (-\frac {1}{n}, b x^{n}\right )}{8 \, n x} + \frac {3 \, \left (-b x^{n}\right )^{\left (\frac {1}{n}\right )} e^{a} \Gamma \left (-\frac {1}{n}, -b x^{n}\right )}{8 \, n x} - \frac {\left (-3 \, b x^{n}\right )^{\left (\frac {1}{n}\right )} e^{\left (3 \, a\right )} \Gamma \left (-\frac {1}{n}, -3 \, b x^{n}\right )}{8 \, n x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b*x^n)^3/x^2,x, algorithm="maxima")

[Out]

1/8*(3*b*x^n)^(1/n)*e^(-3*a)*gamma(-1/n, 3*b*x^n)/(n*x) - 3/8*(b*x^n)^(1/n)*e^(-a)*gamma(-1/n, b*x^n)/(n*x) +
3/8*(-b*x^n)^(1/n)*e^a*gamma(-1/n, -b*x^n)/(n*x) - 1/8*(-3*b*x^n)^(1/n)*e^(3*a)*gamma(-1/n, -3*b*x^n)/(n*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {sinh}\left (a+b\,x^n\right )}^3}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x^n)^3/x^2,x)

[Out]

int(sinh(a + b*x^n)^3/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{3}{\left (a + b x^{n} \right )}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b*x**n)**3/x**2,x)

[Out]

Integral(sinh(a + b*x**n)**3/x**2, x)

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